8.3 Weak Acid and Base Equilibria · 8.4 Acid-Base Reactions and Buffers

8.3 Weak Acid and Base Equilibria · 8.4 Acid-Base Reactions and Buffers

A solution’s pH depends not only on which acid is present, but also on how extensively that acid transfers protons to water. Equal-concentration solutions of hydrochloric acid, carbonic acid, $H_2CO_3$, and acetic acid, $HC_2H_3O_2$, do not produce equal hydronium concentrations because strong acids ionize essentially completely while weak acids ionize only partially.

Weak-acid equilibrium: partial ionization, not incomplete reaction

A weak acid is an acid that transfers only a fraction of its available protons to water. Most particles remain as un-ionized $HA$, while smaller amounts become hydronium, $H_3O^+$, and the conjugate base, $A^-$.

The equilibrium is represented by

HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)

The acid ionization constant, $K_a$, measures the extent of this ionization:

$$ K_a=\frac{[H_3O^+][A^-]}{[HA]} $$

Liquid water is omitted because its concentration is effectively constant. A larger $K_a$ means a greater equilibrium tendency to form $H_3O^+$ and $A^-$, so the acid is stronger. The logarithmic measure is

$$ pK_a=-\log K_a $$

Because of the negative sign, smaller $pK_a$ means stronger acid. The common mistake is to use the initial acid concentration as the final hydronium concentration. For a weak acid, most of the initial acid remains $HA$, so $[H_3O^+]$ is significantly smaller than the initial $[HA]$.

Worked example: percent ionization of a weak acid

Suppose $0.100\ \mathrm{M}$ acetic acid has $K_a=1.8\times10^{-5}$. Let $x$ represent the amount of acid that ionizes.

At equilibrium:

$$ [H_3O^+]=x,\qquad [C_2H_3O_2^-]=x,\qquad [HC_2H_3O_2]=0.100-x $$

Substitution into the $K_a$ expression gives

$$ 1.8\times10^{-5}=\frac{x^2}{0.100-x} $$

Because $K_a$ is small, $x$ is much smaller than $0.100$, so $0.100-x\approx0.100$:

$$ x\approx\sqrt{(1.8\times10^{-5})(0.100)} =1.34\times10^{-3}\ \mathrm{M} $$

Thus,

$$ %\ \mathrm{ionization} =\frac{[H_3O^+]{\mathrm{eq}}}{[HA]{\mathrm{initial}}}\times100 =\frac{1.34\times10^{-3}}{0.100}\times100 =1.34% $$

Only about $1.34%$ of the acid ionizes, even though the solution is acidic. Misconception check: “Weak” does not mean “dilute,” and it does not mean “harmless.” It describes the fraction ionized at equilibrium.

Weak-base equilibrium and conjugate-acid behavior

A weak base accepts protons from water only partially. Its equilibrium is

B(aq)+H_2O(l)\rightleftharpoons HB^+(aq)+OH^-(aq)

The base ionization constant, $K_b$, is

$$ K_b=\frac{[HB^+][OH^-]}{[B]} $$

For a weak base, the equilibrium concentration of $OH^-$ does not equal the initial concentration of $B$. Most base particles remain un-ionized, so only a fraction generates $OH^-$.

$$ pK_b=-\log K_b $$

A larger $K_b$ and smaller $pK_b$ indicate a stronger base. The conjugate acid of a weak base can itself donate a proton:

HB^+(aq)+H_2O(l)\rightleftharpoons B(aq)+H_3O^+(aq)

This explains why an equimolar mixture of a weak base and a strong acid is slightly acidic: the strong acid converts $B$ into $HB^+$, and $HB^+$ then produces some $H_3O^+$.

Quantitative acid-base reactions: first count moles, then apply equilibrium

When acid and base solutions are mixed, the proton-transfer reaction occurs quantitatively before the smaller equilibrium adjustment is considered. The limiting reagent is consumed; any excess reagent remains.

For a weak acid and strong hydroxide,

HA(aq)+OH^-(aq)\rightleftharpoons A^-(aq)+H_2O(l)

The practical sequence is:

1. Calculate initial moles of $HA$ and $OH^-$.
2. Use the $1:1$ stoichiometry to identify the limiting reagent.
3. Subtract consumed moles and determine what remains.
4. Include dilution using the total mixed volume.
5. Apply the appropriate equilibrium model.

Three outcomes for weak acid plus strong base

If $HA$ is in excess, both $HA$ and newly produced $A^-$ remain. The solution is a buffer, a mixture that resists pH change because it contains a weak acid and its conjugate base. Its pH is determined from their mole ratio:

$$ \mathrm{pH}=pK_a+\log\left(\frac{n_{A^-}}{n_{HA}}\right) $$

For example, mixing $0.0200\ \mathrm{mol}$ of $HA$ with $0.0120\ \mathrm{mol}$ of $OH^-$ leaves $0.0080\ \mathrm{mol}$ of $HA$ and produces $0.0120\ \mathrm{mol}$ of $A^-$. If $pK_a=4.76$,

$$ \mathrm{pH}=4.76+\log\left(\frac{0.0120}{0.0080}\right)=4.94 $$

The total volume is unnecessary in this ratio because both species occupy the same final volume. If $OH^-$ is in excess, no buffer approximation applies: calculate

$$ [OH^-]=\frac{n_{OH^-,\mathrm{excess}}}{V_{\mathrm{total}}} $$

If the amounts are equimolar, all $HA$ becomes $A^-$. The resulting solution is slightly basic because $A^-$ hydrolyzes:

A^-(aq)+H_2O(l)\rightleftharpoons HA(aq)+OH^-(aq)

Weak base plus strong acid and buffer capacity

For a weak base and strong acid,

B(aq)+H_3O^+(aq)\rightleftharpoons HB^+(aq)+H_2O(l)

Excess $B$ produces a buffer containing $B$ and $HB^+$. Excess acid is handled from the remaining moles of $H_3O^+$ divided by total volume. Equimolar quantities convert all $B$ into $HB^+$, whose reaction with water makes the solution slightly acidic.

A buffer does not neutralize unlimited added acid or base. If it contains much more $A^-$ than $HA$, it can absorb more added acid than base: added $H_3O^+$ consumes $A^-$, whereas added $OH^-$ consumes the smaller supply of $HA$. Misconception check: buffer capacity depends on the amounts of both components, not merely on having a pH near $pK_a$.

AP skill connections and retrieval check

These problems most directly develop Science Practice 5: Mathematical Routines, especially selecting a logical computational pathway, using stoichiometry, tracking units, and attending to precision. They also require Science Practice 4: Model Analysis when deciding whether a solution is dominated by excess strong reagent, a buffer, or conjugate-ion equilibrium, and Science Practice 6: Argumentation when justifying why a weak-base mixture is acidic or a weak-acid equivalence mixture is basic.

Retrieval check: A mixture contains equal moles of a weak acid $HA$ and $NaOH$. Which major species remains after the quantitative reaction, and why is the final solution basic? Answer: $A^-$ remains, and it reacts with water to produce $OH^-$. A mixture containing equal moles of a weak base $B$ and strong acid instead leaves $HB^+$ and is slightly acidic because $HB^+$ produces $H_3O^+$.